Talk:Subcubic graph number
Where the different values of SCG(n)? And what level of growth in this function? In list of googologisms it is considered as uncomputable number. Is this right? Ikosarakt1 (talk) 11:11, November 9, 2012 (UTC) Even SCG(0) seems hard to compute. We can start with a graph with 1 vertex, which can contain 1 loop. This means there are no more loops in any of the remaining graphs. The second graph can have 2 vertices, with three edges connecting them. The third graph can have 3 vertices, with two edges connecting every pair. The fourth graph can have 4 vertices a,b,c,d, with ab, bc, and cd connected by two edges each, and the remaining three pairs connected by one edge each. It seems that this can go on for a very long time. So even SCG(0) might be very large. The growth rate of SCG(n) is at least the level of ψΩ1 (Ωω) in the fast-growing hierarchy. One of these days I will write a blog post explaining the fast-growing hierarchy up to these large ordinals. I'm pretty sure that SCG(13) will be greater than meameamealokkapoowa oompa, and Chris Bird comes to the same conclusion in his Bowers' Named Numbers article. "Uncomputable number" is a rather vague term, but it is certainly true that SCG(n) is a computable function. You can write a program that runs through all possible acceptable sequences of graphs and keeps track of the longest one.Deedlit11 (talk) 17:31, November 16, 2012 (UTC) Thank you that have explained it to me. I'll put it in the appropriate place. Ikosarakt1 (talk) 09:10, November 17, 2012 (UTC) Whoops, I messed up the SCG(0) analysis. I'm now quite certain that SCG(0) = 6. The first graph is a single vertex with a loop; this prevents all loops or cycles in future graphs. The second graph is two vertices with a single edge between them. This prevents all edges of any kind in future graphs. So the remaining graphs are all edgeless graphs. So the last four graphs are graphs with 3, 2, 1, and 0 vertices with no edges between them. So SCG(0) = 6, or 5 if the empty graph is not allowed. SCG(1), on the other hand, is provably very large. We can construct a sequence of graphs (for convenience I will start with G2, so that Gn has n vertices.) G2 = 2 vertices with three edges between them. G3 = vertices a, b, and c with two edges between a and b and one edge between b and c. G4 = two pairs of vertices each with two edges between them. G5 = a 5-cycle, i.e. the graph of a pentagon. G6 = a four-cycle (square) plus two vertices with an edge between them. G7 = a square plus three vertices each with a loop. G8 = a square plus two looped vertices and two unlooped vertices. G9 = a square plus two looped vertices and one unlooped vertex. G10 = a square plus two looped vertices. G11 = a square plus one looped vertex and six unlooped vertices. G12 = a square plus one looped vertex and five unlooped vertices. ... G17 = a square plus one looped vertex. G18 = a square plus 14 unlooped vertices. G19 = a square plus 13 unlooped vertices. ... G32 = a square. We can then define G33 to be a triangle plus a 30-vertex cubic tree. We can then continue with a triangle plus a very long sequence of cubic forests. Let forest(n) be the length of the longest sequence of cubic forests such that the ith forest has at most n + i vertices. Then we can continue our above sequence with a sequence of length forest(29) consisting of a triangle plus a cycleless cubic forest. We can then continue with a sequence of length forest(forest(29)) consisthing of 2-cycle plus a cyclelss cubic forest. Then we construct a sequence of length forest(forest(forest(29))) consisting for a looped vertex plus a cycleless cubic forest. We finish with a sequence of length forest(forest(forest(forest(29)))) consisting of cycleless cubic forests.' I believe the growth rate of forest(n) is about epsilon_0 in the fast-growing hierarchy, SCG(1) > forest(forest(forest(forest(29)))) is very large, clearly indeterminable.Deedlit11 (talk) 03:08, November 23, 2012 (UTC) According to you, SCG(1) is larger than goppatothplex, but it is a mere lower bound... Ikosarakt1 (talk) 12:15, November 26, 2012 (UTC) I just had an idea of how this sequence can be made a lot longer - if I'm not mistaken non-simple subcubic graphs can have loop on leaf vertex (if they can't, you can stop reading here). Define G'3 as Deedlit's G3 with loop around vertex c. G'4 is same as G3 and G'5 is same as G4. Then G'6 is hexagon graph. G'7 will be pentagon with two vertices with loops and edge between them. G'9 will be just G6 (I haven't made any analysis, so G'8 and G'9 just delete leaf loops, for simplicity). Continuing in same manner as Deedlit we'll at some point reach alone pentagon. Say it happens at G'x. Using cubic trees we can now reach forest^5(x-4). We can also insert stages between triangle plus a cycleless forest and 2-cycle plus a cycleless forest by using 2 vertices with loops around them, and then 2 vertices from which only one has loop. This creates forest^7(x-4) graphs long sequence. And for the end another lengthening - suppose that after using forests with 2-cycles we make forests and 3 looped vertices, then 2 looped vertices, then one. With some optimization we can create sequences reaching forest^(forest(x-4))(x-4) or even ones with function exponent having function exponent! LittlePeng9 (talk) 18:02, February 5, 2013 (UTC) Yeah, I wasn't trying to maximize the sequence because it was taking too much explaining to do. I would be interested to see how high you can go! Deedlit11 (talk) 04:55, March 7, 2013 (UTC) I created whole blog post defining and explaining 3 functions I later used to bound SCG(1) from below. Can you come up with better result? LittlePeng9 (talk) 20:24, March 10, 2013 (UTC) Rayo's number How does SCG(13) compare to Rayo's number? FB100Z • talk • 02:52, January 2, 2013 (UTC) Rayo's function grows as fast as \(f_{\omega_{1}^{CK} \times \omega}(n)\) (I found it at cp4space). Busy beaver function is only \(f_{\omega_{1}^{CK}}(n)\). BB(n) is yet uncomputable, and Ra(n) all the more so. SCG(n), however, is computable, so Rayo's number should be larger. Ikosarakt1 (talk) 16:49, January 19, 2013 (UTC) Look on this CP4space again. Together with A.P.Goucher we concluded that Rayo's function strength is at level of \(f_{\omega_{\omega}^{CK}}(n)\) (ordinal in there is smallest limit of admissibles, obviously still countable) LittlePeng9 (talk) 16:18, January 28, 2013 (UTC) Actual strength of SCG function Deedlit states that strength of SCG function is at least \(\psi_{\Omega_1}(\Omega_\omega)\). I guess he says so, because it is provably strength of Robertson-Seymour theorem. But this theorem is saying about well-quasi-ordering set of ''all ''graphs. Subcubic graphs make only tiny fraction of whole. I think strength of SCG function is strictly smaller than this ordinal. What do you think about that? On the other hand, at the end of Wikipedia article on graph minor theorem there is finite form concerning every possible graph with graph minorship relation. Function based on this finite form has at least \(\psi_{\Omega_1}(\Omega_\omega)\) as corresponding ordinal, and I'm almost sure I can prove that LittlePeng9 (talk) 20:42, February 13, 2013 (UTC) Actually, I wasn't basing the strength of the SCG function on the Robertson-Seymour theorem, but rather on Friedman's statements in his FOM postings. In particular, he statest that the fact that finite subcubic graphs are WQO under homeomorphic embeddability is not provable in the theory Pi-1-1-CA_0. This implies that SCG(n) is not a provably recursive function of Pi-1-1-CA_0, and the proof-theoretic ordinal of pi-1-1-CA_0 is \(\psi_{\Omega_1}(\Omega_\omega)\). In fact Friedman states that even the fact that finite planar subcubic graphs (and presumably finite planar simple subcubic graphs) are WQO under homeomorphic emeddability is not provable in Pi-1-1-CA_0, so PSCG(n) and PSSCG(n) won't be provably recursive in Pi-1-1-CA_0. Now, we may presume that a function is provably recursive in Pi-1-1-CA_0 if and only if it is elementary recursive in some function \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\). So a function not provably recursive in Pi-1-1-CA_0 will not be elementary recursive in any \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\). However, this does not mean that the function will dominate all \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\); for example, it could alternate between large values and 0. (Or, for an example of an increasing function, it could alternate between periods of growing very fast with growing very slowly, with the periods getting longer and longer.) So perhaps we cannot state with certainty that the previous functions grow at a certain rate. However, it seems highly unlikely to me that any natural function will have such strange behavior; so I am almost certain that PSSCG(n) et al grow at \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) or faster. Also, Friedman states as a theorem that SCG(13) is larger than the halting time of any Turing machine starting from the blank tape that can be proven to halt in at most \(2 \uparrow \uparrow 2000 \) symbols, based on the fact that \(SCG(13) \ge PSCG(2 \uparrow \uparrow 2000) \). So somehow he knows that \(PSCG(n) > f(n) \), where f(n) is the halting time of the longest running Turing machine starting from the blank tape that can be proven to halt in at most n symbols. I think it is clear that the latter grows at the rate of \(\psi_{\Omega_1}(\Omega_\omega)\) or higher. How Friedman knows this, I don't know, but I will trust him. Deedlit11 (talk) 04:50, March 7, 2013 (UTC) I wonder how the values of SCG(n) compare to numbers derived from the Buchholz hydra game? --Ixfd64 (talk) 18:35, March 7, 2013 (UTC) Good question! If we let BHydra(n) be the time it takes to slay a Buchholz hydra consisting of an (n+2)-path labelled with n n's (the first two vertices are required to be labelled with + and 0), then both SCG(n) and BHydra(n) are about at the level of \(\psi_{\Omega_1}(\Omega_\omega)\). BHydra(n) will in fact be roughly equivalent to \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) , as the structure of Buchholz Hydras with finite ordinal labels mimic ordinal notations up to \(\psi_{\Omega_1}(\Omega_\omega)\), and the act of slaying the hydras mimics taking fundamental sequences. On the other hand, from Friedman's work we can infer that \(PSCG(n) \ge F_{\psi_{\Omega_1}(\Omega_\omega)} \), and SCG(n) grows faster than PSCG(n) (for example \(SCG(13) > PSCG(2\uparrow\uparrow 2000) \). So I believe SCG(n) outpaces BHydra(n). However, if we label Buccholz hydras with infinite ordinals, than I bet that the numbers derived from Buchholz hydras will grow faster. Deedlit11 (talk) 00:33, March 8, 2013 (UTC) From Buchholz's work we know that for every hydra we can prove its termination in \(\Pi^1_1-CA_0\), but we can't do this uniformly (otherwise, Theorem III from this paper could be disproven by induction on number of labels). So we know that function F(n) related to (n+2)-path labelled with n ω's is unprovably recursive in \(\Pi^1_1-CA_0\), so it almost certainly outbounds \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) LittlePeng9 (talk) 10:25, March 9, 2013 (UTC) Blog entry about SCG function Someone wrote an interesting blog entry on the SCG function here. I suspect it's someone from this wiki. --Ixfd64 (talk) 20:09, June 26, 2013 (UTC) :Wondering who is the author? Konkhra (talk) 21:27, June 26, 2013 (UTC). :When the author talks about increasing the valence limit of SCG, it would be easier to go to Minor(n) instead. We can define \(Minor_f\) as the longest sequence of graphs \(G_n\) such that the number of vertices of \(G_n\) is no more than \(f(n)\) and for no i < j do we have \(G_i\) a minor of \(G_j\). Note that it is important that we go from topological minor to regular minor for valence greater than 3, otherwise we do not necessarily get a finite sequence. :The array extension at the end just adds \(\omega^2\) to the ordinal, so it's pretty unimportant. Deedlit11 (talk) 23:48, June 26, 2013 (UTC) I think this extension can add \(\omega^\omega\) to the order type, as it works similar to Bowers' variant of array notation. Ikosarakt1 (talk ^ ) 19:57, June 29, 2013 (UTC) :But it works more like Conway chained arrow notation, since the recursion rule decrements the highest entry rather than the lowest nonzero entry. Deedlit11 (talk) 00:17, June 30, 2013 (UTC) 2^1000 I have heard about a^b as a notation for \(a \uparrow\uparrow b\). Is it true? Ikosarakt1 (talk) 17:38, March 1, 2013 (UTC) :Yes, really. Harvey Friedman wrote: "Here 2 ^ 1000 is an exponential stack of 2's of height 1000" Here's the link: http://www.cs.nyu.edu/pipermail/fom/2006-March/010244.htmlKonkhra (talk) 09:07, March 2, 2013 (UTC) Informal explanation? Is there a way to explain SCG less esoterically, in a way similar to TREE sequence#Explanation? It'd be nice if we could find a way to encode subcubic graphs in strings so that the concept of a graph minor is really clear. FB100Z • talk • 19:12, March 17, 2013 (UTC) I don't think so - I think best we can do is explain what graph and topological minors are, and how it connects to homeomorphic embeddability. LittlePeng9 (talk) 11:00, March 18, 2013 (UTC) I don't know of any natural interpretation of graphs as strings, other than to list out the adjacency matrix, and I don't it will be easy to define topological minors using that.Deedlit11 (talk) 12:49, March 18, 2013 (UTC) extending the function? Is it possible to generalize or extend the SCG function to make it more powerful? Just curious. --Ixfd64 (talk) 05:17, April 17, 2013 (UTC) There is certainly no obvious way to escape corresponding ordinal of SCG function. Giving more freedom for size of graphs won't increase asymptotic growth, and throwing subcubicity away may break totality of function (infinite sequences). We can allow general simple graphs if we consider graph (instead of topological) minor relation, which is WQO over set of all graphs. But it still doesn't have larger ordinal. LittlePeng9 (talk) 05:31, April 17, 2013 (UTC) I'll use Minor(n) for Robertson-Seymour function. While SCG(n) and Minor(n) functions have the same minimum ordinal, neither has a good maximum ordinal (except perhaps the proof-theoretic ordinal of a much stronger theory), and it's quite possible that the ordinal for Minor(n) is larger than the ordinal for SCG(n). Deedlit11 (talk) 06:49, April 17, 2013 (UTC) :What is the function Minor(n)? Where I can read about it? Konkhra (talk) 12:51, May 24, 2013 (UTC) ::Wonder how large Minor(1) or Minor(2)? Konkhra (talk) 03:14, May 25, 2013 (UTC) ::Minor(n) is length of longest sequence of simple graphs G such that Gi has at most i+n vertices and no Gi is graph minor of Gj for i=SSCG(n), I also guess Minor(2)=SSCG(2). LittlePeng9 (talk) 08:28, May 25, 2013 (UTC) growth rate of SCG(k) upper-bounded by TFB ordinal? This paper suggests that the strength of the graph minor theorem is limited by \(\Pi_1^1\)-\(\text{CA}\)+\(\text{BI}\). I'm not sure of the difference between \(\Pi_1^1\)-\(\text{CA}\)+\(\text{BI}\) and , but does this mean the ordinal of the SCG function is limited by the Takeuti-Feferman-Buchholz ordinal? --Ixfd64 (talk) 16:50, June 27, 2013 (UTC) As far as I know, subscript in \(\text{CA}_0\) is often dropped, so both mentioned systems are equal. Friedman's result gives lower bound corresponding to system \(\Pi_1^1-CA_0\), which excludes so called bar induction. Paper you linked doesn't prove what you said; that was already proven by Robertson and Seymour. I think that result shows that ordinal of SCG function is at most TFB ordinal (with tiny chance for equality). LittlePeng9 (talk) 20:54, June 27, 2013 (UTC) :The zero subscript definitely makes a difference, although I forget exactly what it meant. (It affects the scope of the comprehension axiom.) Since \(\Pi_1^1\)-\(\text{CA}\)+\(\text{BI}\) has the TFB ordinal as its proof-theoretic ordinal, this result shows that SCG(n) and Minor(n) indeed are at most the TFB ordinal in the FGH. Deedlit11 (talk) 03:16, June 28, 2013 (UTC) :Indeed, zero subscript means restricted induction. In this case, comprehension, so also induction, is limited to \(\Pi^1_1\) formulas. But this system without inductive restriction is equivalent to full second order arithmetic. What I meant is that subscript is often ommited in this particular system, though I may be completely wrong. LittlePeng9 (talk) 06:08, June 28, 2013 (UTC) ::I don't believe systems with the full second order induction scheme are equivalent to second order arithmetic; for example, I have read that the proof theoretic ordinal of ACA is \(\epsilon_{\epsilon_0}\), as opposed to ACA_0 which has proof theoretic ordinal \(\epsilon_0\). So ACA is strictly between ACA_0 and second order arithmetic. I believe the same thing is true of \(\Pi_1^1\)-\(\text{CA}\) versus \(\Pi_1^1\)-\(\text{CA}_0\) The paper that Ixfd64 linked suggests they are different as well. Deedlit11 (talk) 06:31, June 28, 2013 (UTC) ::It might be, I just said what I thought is true. Everyone is allowed to be mistaken. Do you maybe know where is this result about ACA mentioned? LittlePeng9 (talk) 16:28, June 28, 2013 (UTC) :::Certainly everyone is allowed to be mistaken, no judgment here! Not sure where I read it, but a quick Google search yields the following: http://mathforum.org/kb/message.jspa?messageID=5973538 Deedlit11 (talk) 16:56, June 28, 2013 (UTC) I just found a presentation from the University of Wisconsin that says the order type of the graph minor theorem is greater than the TFB ordinal. However, all the other papers and discussions I've read say otherwise. Would it be safe to assume that the UW paper is incorrect? --Ixfd64 (talk) 00:32, July 2, 2013 (UTC) :I believe he is mistaken, because he says that the proof theoretic ordinal of \(\Pi_1^1\)-\(\text{CA}_0\) is \(\psi_0 (\varepsilon_{\Omega_{\omega}+1})\), whereas I have read from multiple sources that the proof theoretic ordinal is \(\Pi_1^1\)-\(\text{CA}_0\) is \(\psi_0 (\Omega_{\omega})\). Nice slides though, please keep sharing your finds. Deedlit11 (talk) 04:18, July 2, 2013 (UTC) It's pretty funny how two papers you linked are stating contrary results about how RS theorem relates to TFB ordinal. This slideshow is also nicely introducing wqo's. This is I think where I first heard of them. LittlePeng9 (talk) 08:45, July 2, 2013 (UTC) Values of SSCG I found: SSCG(0) = 2 SSCG(1) = 5, using the following sequence: •–• ••• •• • and the empty graph Also, is an empty graph allowed? Wythagoras (talk) 09:16, June 28, 2013 (UTC) :It's not clear what Friedman intended, but I have been allowing it. (The article currently says SCG(0) = 6 which assumes the empty graph is allowed.) Of course, it just makes the numbers one greater. :You may be interested in the following article http://cp4space.wordpress.com/2013/01/13/graph-minors/, which among other things discusses the values of SSCG(2) and SSCG(3). Note that SSCG(3) is _extremely_ large! Deedlit11 (talk) 09:34, June 28, 2013 (UTC) :Can we find values for PSCG(n) and PSSCG(n)? And for SSCG(3), what does very large n mean? Can it be larger than TREE(3) itself? or even TREE^(TREE(3))(3)? Wythagoras (talk) 10:42, June 28, 2013 (UTC) ::Regarding PSCG(n) and PSSCG(n), note that a graph is planar if and only if it does not contain a subdivision of K_5 or K_3,3 as a subgraph. (This is Kurotowski's Theorem, which is usually used for simple graphs but one can see that it applies to multigraphs as well.) Since K_5 is a degree 4 graph, a subcubic graph will never contain a subdivision of K_5, so a subcubic graph is planar if and only if it doesn't contain a subdivision of K_3,3, which has 6 vertices. So PSSCG(n) = SSCG(n) for n < 5. However PSCG(n) < SCG(n) and PSSCG(n) < SSCG(n) for n >= 6; it is a reasonable conjecture that SSCG(5) = PSSCG(6). Deedlit11 (talk) 14:59, June 28, 2013 (UTC) :::How we could know that a K_3,3 cannot appear on the expansion of SCG(5)? The answer will help proving that PSCG(n) = SCG(n) for n < 6. -- ☁ I want more ⛅ 13:39, June 30, 2013 (UTC) :::We can check by cases that every 5-vertex subcubic graph is minor of K3,3. We can check trees, graphs with 3-cycle but not 4-cycle, with 4-cycle but not 5-cycle and with 5-cycle. First two are easy, latter two may have fused cycles, so may be a little problem, but it isn't that hard. LittlePeng9 (talk) 15:09, June 30, 2013 (UTC) :::No, wait, it works for simple graphs only! If we take a triangle with one edge with a loop added, then triangle with single edge added and separate loop, next graph can be K3,3! Also, it doesn't work for SSCG(5), because then K3,3 can be first graph. LittlePeng9 (talk) 15:30, June 30, 2013 (UTC) ::::You're both right; I corrected the post. So, by LittlePeng9's example, PSCG(3) < SCG(3). Obviously PSCG(0) = SCG(0). What about PSCG(1) and PSCG(2)? Similar questions for PSSCG(n) versus SSCG(n): Clearly PSSCG(n) = SSCG(n) for n < 4, but what about PSSCG(4)? Deedlit11 (talk) 17:32, June 30, 2013 (UTC) :::::About PSCG(2) we have following sequence: path of length 2 (i.e. with 2 edges) and loops on both ends. Then K1,3 (called claw or star) with loop on one vertex. Now path of length 5 with loop on one end. K3,3 is now valid graph, butI doubt it can lead to very long sequence. I think PSCG(1)=SCG(1), unless we can find graphs on 3, 4 and 5 vertices, each with exactly one cycle (loop or double edge, to be exact), none of which is minor of other or K3,3. I also gave reason why PSSCG(4)=SSCG(4) (it must start with 5-vertex graph whichis minor of K3,3). LittlePeng9 (talk) 17:55, June 30, 2013 (UTC) ::::::Hmmm, now that I think about it, we don't know that PSCG(3) < SCG(3) unless we know that the starting sequence for SCG(3) is optimal, which could be very hard to prove. I've found a sequence for SCG(1) that leads to K_3,3: start with an edge with loops at both ends. Next is an edge with a loop at one end, plus a loop. Next is four loops, followed by three loops, followed by K_3,3. But we don't know that this leads to a longer sequence than PSCG(1) unless we can prove that the longest sequence for PSCG(1) starts the same way. I've made a ton of mistakes here! Deedlit11 (talk) 18:07, June 30, 2013 (UTC) ::::::I also just found nonplanar sequence for SCG(1). I wanted to ask, is your bound SSCG(7)>PSSCG_PSSCG(n)(n) about SSCG(6) or SSCG(7)? I tried to do it for SSCG(6), but I couldn't do much with K3,3 with one added vertex. LittlePeng9 (talk) 18:22, June 30, 2013 (UTC) ::::::It's about SSCG(7). I too couldn't do much with just one added vertex, but with two added vertices you can go on for a LONG time. Deedlit11 (talk) 18:25, June 30, 2013 (UTC) ::SSCG(3) is indeed larger than TREE^(TREE(3))(3). If we define TREE_2 = TREE^n (n), TREE_3 = (TREE_2)^n (n), and TREE_4 = (TREE_3)^n (n), then SSCG(3) is at least TREE_4 (n) for n large (I haven't worked out how large, but clearly greater than Graham's number for example.). Deedlit11 (talk) 15:05, June 28, 2013 (UTC) ::Thanks. And SSCG(4)? is it bigger then TREE_SSCG(3)(SSCG(3)) for example? Wythagoras (talk) 08:04, June 29, 2013 (UTC) :::I must admit I haven't thought much about SSCG(4). It might be bigger than TREE_SSCG(3) (SSCG(3)), but it would be hard to prove it, as we would need to employ a sequence of SSCG(3) graphs somewhere, and that would leave just one extra vertex to use, which doesn't seem like enough. :::I can prove that SSCG(7) > PSSCG_PSSCG(n)(n) for a large n, so SSCG grows much faster than PSSCG, and similarly SCG grows much faster than PSCG. Deedlit11 (talk) 19:20, June 29, 2013 (UTC) Also, I believe I can prove that SSCG(3n+4) >= 3 SCG(n), improving Goucher's inequality. We can convert any multigraph into a simple graph by the following procedure: Whenever we have multiple edges between two vertices, we subdivide all but one of the edges into two edges. Whenever we have a loop, we subdivide it into three edges. Each vertex either has a loop, a double or triple edge, or neither. In any case we need to add at most two vertices to the adjoining edges. So this conversion at most triples the number of edges. So, let's say we have a sequence of SCG(n) multigraphs with at most n+1, n+2, ... n+SCG(n) vertices. The conversion converts this to a sequence of simple graphs with at most 3n+3, 3n+6, ... 3n + 3 SCG(n) vertices. Since each subsequent graph in the sequence has the max number of vertices increasing by 3, we replace each graph by a sequence of three graphs, by replacing G by G + 2 vertices, G + 1 vertex, G. This leads to a sequence with 3n+5, 3n+4, 3n+3, 3n+8, 3n+7, 3n+6,... vertices. This is a valid sequence for SSCG(3n+4) of length 3SCG(n), so SSCG(3n+4) >= 3SCG(n). Deedlit11 (talk) 16:12, June 28, 2013 (UTC) I found that SSCG(6)>=PSSCG(12)+6 and SCG(6)>=PSCG(7+PSCG(1))+PSCG(1)+1. I believe first equality holds, but I doubt about second one. As we are not even sure if SCG(3)=PSCG(3) (I think we can prove it for 1 and 2 by checking all sequences leading to K3,3) sequence I used may be suboptimal. LittlePeng9 (talk) 15:22, July 1, 2013 (UTC) :Yes, I agree with both inequalities. Although we might as well say SCG(6)>=PSCG(7+SCG(1))+SCG(1)+1 until we can actually prove that SCG(1) = PSCG(1). Deedlit11 (talk) 15:38, July 1, 2013 (UTC) Another variant Consider SSCG function with following restriction - every graph must consist of one connected component (or is empty graph). I checked that CSSCG(0)=2, CSSCG(1)=4, CSSCG(2)=8. After many tries I was able to prove CSSCG(3)>TREE(3), and heurestically that CSSCG(3)>TREE(5), though it'd takeme a while to verify. The key was Goucher's reduction, which thankfully results in connected graphs. I noticed CSCG(1) won't be really big number, though it may reach epsilon_0. LittlePeng9 (talk) 10:39, July 2, 2013 (UTC) Grows CSSCG faster then PSSCG? Wythagoras (talk) 13:22, July 2, 2013 (UTC) I believe connected variant grows strictly slower than planar one. This is only intuition, though. LittlePeng9 (talk) 14:15, July 2, 2013 (UTC) :I get that CSSCG(1) = 3; edge, point, empty graph. I agree that CSSCG(2) = 8. As for CSSCG(3), we can improve Goucher's reduction to TREE(n) a little bit: instead of having the label connect to a triangle, we can have the label connect directly to the main cycle. We can still distinguish the label since it will be an acyclic tree, whereas the children nodes will always have a cycle. This will allow the sequence to have fused cycles with triangles in addition to the fused cycles with squares that Goucher has, before going to the TREE sequence. Anyway, I'd be interested to hear you analysis for CSSCG(3) > TREE(3) and CSSCG(3) > TREE(5) heuristically. Deedlit11 (talk) 18:46, July 2, 2013 (UTC) ::Yes, CSSCG(1)=3, for some reason I counted loop graph. But your improvement doesn't work - if a label is connected just as child nodes are (which is a case in your method) then tree can interfere with ancestors of node. Simplest example: suppose we have one node tree with label represented by path of length, say, 4. In the next tree root very likely has a child, but because label was connected just as child is, the second node can't have 4-path as a minor, so it restricts more than TREE function does. I'll post my analysis tomorrow, as soon as I find my notes, but I think I actually failed with TREE(5) bound. LittlePeng9 (talk) 21:10, July 2, 2013 (UTC) ::I found my notes (A4 sheet of drawing paper filled with graphs of all kinds) and here is sequence of graphs I found: ::G4: K4 ::G5: K2,3 ::G6: square with two triangles fused ::G7: square and pentagon fused. Let R denote two fused squares, with long and short edges defined intuitively. ::G8: R with single points attached two corners on longer edge ::G9: R with 1-path attached to one vertex and single point attached to opposite vertex ::G10: same as above, but with 1-path reduced to point ::Let Ra,b denote R with a vertices added to one vertex, and b vertices added to adjacent one. Then G11-24 are: R5,0 R4,2 R4,1 R4,0 R3,3 R3,2 R3,1 R3,0 R2,2 R2,1 R2,0 R1,1 R1,0 R. While writing this I noticed I can use trees too, so there are big hopes for much more than TREE(5)! ::Now with G25 we start TREE simulation with tree with counter 10 and single node with any 4-vertex label. G35 is same with counter of length 0, and then we continue. We have two 3-vertex trees which we can use as labels, but it's enough for TREE(3), as we use first label only once. So CSSCG(3)>TREE(3), Q.E.D. LittlePeng9 (talk) 08:40, July 3, 2013 (UTC)